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ISSN 1364-0380 (on line) 1465-3060 (printed) 331

Geometry & Topology GGGGGGGGG GG

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T TTTTTT

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Volume 3 (1999) 331–367

Published: 14 October 1999

Examples of Riemannian manifolds with positivecurvature almost everywhere

Peter Petersen

Frederick Wilhelm

Department of Mathematics, University of CaliforniaLos Angeles, CA 90095, USA

andDepartment of Mathematics,University of California

Riverside, CA 92521-0135, USA

Email: [email protected] and [email protected]

Abstract

We show that the unit tangent bundle of S4 and a real cohomology CP 3 admitRiemannian metrics with positive sectional curvature almost everywhere. Theseare the only examples so far with positive curvature almost everywhere that arenot also known to admit positive curvature.

AMS Classification numbers Primary: 53C20

Secondary: 53C20, 58B20, 58G30

Keywords: Positive curvature, unit tangent bundle of S4

Proposed: Steve Ferry Received: 27 March 1999

Seconded: Gang Tian, Walter Neumann Revised: 30 July 1999

c© Geometry & Topology Publications

332 Peter Petersen and Frederick Wilhelm

0 Introduction

A manifold is said to have quasi-positive curvature if the curvature is nonneg-ative everywhere and positive at a point. In analogy with Aubin’s theoremfor manifolds with quasi-positive Ricci curvature one can use the Ricci flowto show that any manifold with quasi-positive scalar curvature or curvatureoperator can be deformed to have positive curvature.

By contrast no such result is known for sectional curvature. In fact, we donot know whether manifolds with quasi-positive sectional curvature can be de-formed to ones with positive curvature almost everywhere, nor is it knownwhether manifolds with positive sectional curvature almost everywhere can bedeformed to have positive curvature. What is more, there are only two ex-amples ([12] and [11]) of manifolds with quasi-positive curvature that are notalso known to admit positive sectional curvature. The example in [11] is a fakequaternionic flag manifold of dimension 12. It is not known if it has positivecurvature almost everywhere. The example in [12] is on one of Milnor’s exotic7-spheres. It was asserted without proof, in [12], that the Gromoll–Meyer met-ric has positive sectional curvature almost everywhere, but this assertion wasdisproven by Mandell, a student of Gromoll ([16], cf also [25]). Thus there is noknown example of a manifold with positive sectional curvature at almost everypoint that is not also known to admit positive curvature. We will rectify thissituation here by proving the following theorem.

Theorem A The unit tangent bundle of S4 admits a metric with positivesectional curvature at almost every point with the following properties.

(i) The connected component of the identity of the isometry group is iso-morphic to SO(4) and contains a free S3–subaction.

(ii) The set of points where there are 0 sectional curvatures contains totallygeodesic flat 2–tori and is the union of two copies of S3×S3 that intersectalong a S2 × S3 .

Remark In the course of our proof we will also obtain a precise description ofthe set of 0–curvature planes in the Grassmannian. This set is not extremelycomplicated, but the authors have not thought of a description that is succinctenough to include in the introduction.

Remark In the sequel to this paper, [25], the second author shows that themetric on the Gromoll–Meyer sphere can be perturbed to one that has posi-tive sectional curvature almost everywhere. In contrast to [25] the metric weconstruct here is a perturbation of a metric that has zero curvatures at everypoint.

Geometry & Topology, Volume 3 (1999)

Manifolds with positive curvature almost everywhere 333

By taking a circle subgroup of the free S3–action in Theorem A(i) we get thefollowing.

Corollary B There is a manifold M6 with the homology of CP 3 but not thecohomology ring of CP 3 , that admits a metric with positive sectional curvaturealmost everywhere.

There are also flat totally geodesic 2–tori in M6 so as a corollary of Lemma4.1 in [22] (cf also Proposition 3 in [5]) we have the following.

Corollary C There are no perturbations of our metrics on the unit tangentbundle and M6 whose sectional curvature is positive to first order.

That is, there is no smooth family of metrics {gt}t∈R with g0 the metric inTheorem A or Corollary B so that

d

dtsecgt(P )|t=0 > 0

for all planes P that satisfy secg0(P ) = 0.

Remark Although the unit tangent bundle of S4 is a homogeneous space (seebelow), the metric of Theorem A is obviously inhomogeneous. What is more ifthe unit tangent bundle of S4 admits a metric with positive sectional curvature,then it must be for some inhomogenous metric see [6]. The space in CorollaryB is a biquotient of Sp(2). It follows from [19, Theorem 6] that M6 does nothave the homotopy type of a homogeneous space.

Before outlining the construction of our metric we recall that the S3–bundlesover S4 are classified by Z ⊕ Z as follows ([14], [21]). The bundle that corre-sponds to (n,m) ∈ Z ⊕ Z is obtained by gluing two copies of R

4 × S3 togethervia the diffeomorphism gn,m: (R4\{0}) × S3 −→ (R4\{0}) × S3 given by

gm,n(u, v) −→ (u

|u|2 ,umvun

|u|n+m), (0.1)

where we have identified R4 with H and S3 with {v ∈ H | |v| = 1}. We will

call the bundle obtained from gm,n “the bundle of type (m,n)”, and we willdenote it by Em,n .

Translating Theorem 9.5 on page 99 of [15] into our classification scheme (0.1)shows that the unit tangent bundle is of type (1, 1). We will show it is also thequotient of the S3–action on Sp(2) given by

A2,0( p,

(

a bc d

)

) =

(

pa pbpc pd

)

.



(It was shown in [20] that this quotient is also the total space of the bundle oftype (2, 0), so we will call it E2,0 .)

The quotient of the biinvariant metric via A2,0 is a normal homogeneous spacewith nonnegative, but not positive sectional curvature. To get the metric ofTheorem A we use the method described in [8] to perturb the biinvariant ofSp(2) using the commuting S3–actions

Au( p1,

(

a bc d

)

) =

(

p1a p1bc d

)

Ad( p2,

(

a bc d

)

) =

(

a bp2c p2d

)

Al( q1,

(

a bc d

)

) =

(

aq1 bcq1 d

)

Ar( p1,

(

a bc d

)

) =

(

a bq2

c dq2

)

.

We call the new metric on Sp(2), gν1,ν2,lu1,ld1

, and will observe in Proposi-tion 1.14 that A2,0 is by isometries with respect to gν1,ν2,lu

1,ld1

. Our metricon the unit tangent bundle is the one induced by the Riemannian submersion

(Sp(2), gν1,ν2,lu1,ld1

)q2,0−→ Sp(2)/A2,0 = E2,0 .

In section 1 we review some generalities of Cheeger’s method. In section 2we study the symmetries of E2,0 . In section 3 we analyze the infinitesimal

geometry of the Riemannian submersion Sp(2)p2,1−→ S7 , given by projection

onto the first column. This will allow us to compute the curvature tensor ofthe metric, gν1,ν2

, obtained by perturbing the biinvariant metric via Al andAr . In section 4 we compute the A–tensor of the Hopf fibration S7 −→ S4 ,because it is the key to the geometry of gν1,ν2

. In section 5 we specify thezero curvatures of gν1,ν2

, and in section 6 we describe the horizontal space ofq2,0: Sp(2) −→ E2,0 with respect to gν1,ν2

and hence (via results from section1) with respect to gν1,ν2,lu

1,ld1

. In section 7 we specify the zero curvatures ofE2,0 , first with respect to gν1,ν2

and then with respect to gν1,ν2,lu1,ld1

, provingTheorem A. In section 8 we establish the various topological assertions thatwe made above, that E2,0 is the total space of the unit tangent bundle andthat while the cohomology modules of E2,0/S

1 are the same as CP 3 ’s the ringstructure is different. Using these computations we will conclude that E2,0 andE2,0/S

1 do not have the homotopy type of any known example of a manifoldof positive curvature.

We assume that the reader has a working knowledge of O’Neill’s “fundamentalequations of a submersion” [18] and the second author’s description of the



tangent bundle of Sp(2), [24], we will adopt results and notation from bothpapers, in most cases without further notice. It will also be important for thereader to keep the definitions of the five S3 actions Au , Ad , Al , Ar and A2,0

straight. To assist we point out that the letters u, d, l and r stand for “up”,“down”, “left” and “right” and are meant to indicate the row or column ofSp(2) that is acted upon.

Acknowledgments We are grateful to Claude Lebrun, Wolfgang Ziller, andthe referee for several thoughtful and constructive criticisms of the first draftof this paper.

The first author is supported in part by the NSF. Support from National Sci-ence Foundation grant DMS-9803258 is gratefully acknowledged by the secondauthor.

Dedicated to Detlef Gromoll on his sixtieth birthday

1 Cheeger’s Method

In [8] a general method for perturbing the metric on a manifold, M , of non-negative sectional curvature is proposed. Various special cases of this methodwere first studied in [4] and [7].

If G is a compact group of isometries of M , then we let G act on G × M by

g(p,m) = (pg−1, gm). (1.1)

If we endow G with a biinvariant metric and G × M with the product metric,then the quotient of (1.1) is a new metric of nonnegative sectional curvature onM . It was observed in [8], that we may expect the new metric to have fewer 0curvatures and symmetries than the original metric.

In this section we will describe the effect of certain Cheeger perturbations onthe curvature tensor of M . Most of the results are special cases of results of[8], we have included them because they can be described fairly succinctly andare central to all of our subsequent computations.

The quotient map for the action (1.1) is

qG×M : (p,m) 7→ pm. (1.2)

The vertical space for qG×M at (p,m) is

VqG×M= {(−k, k) | k ∈ g}



where the −k in the first factor stands for the value at p of the killing field onG given by the circle action

(exp(tk), p) 7→ p exp(−kt) (1.3)

and the k in the second factor is the value of the killing field

d

dtexp(tk)m (1.4)

on M at m.

Until further notice all Cheeger perturbations under consideration will have theproperty:

For all k1, k2 ∈ g if 〈(k1, 0), (k2, 0)〉 = 0, then 〈(0, k1), (0, k2)〉 = 0, (1.5)

where g is the Lie algebra of G. Notice that Au , Ad , Ar and Al have thisproperty.

In this case the horizontal space for qG×M is the direct sum

HqG×M= {(λ2

2k, λ21k) |k ∈ g} ⊕ ( {0} × HOG

), (1.6)

where λ1 = |(−k, 0)| and λ2 = |(0, k)| and HOGis the space that is normal to

the orbit of G. The image of (λ22k, λ2

1k) under dqG×M is

dqG×M (λ22k, λ2

1k) =d

dtp exp(λ2

2kt) exp(λ21kt)m|t=0 = dLp,∗( (λ2

1 + λ22)k ). (1.7)

It follows that the effect of Cheeger’s perturbation is to keep HOGperpendicular

to the orbits of G, to keep the metric restricted to HOGunchanged and to

multiply the length of the vector dLp,∗( k ) by the factor

√

(λ22λ1)2 + (λ2

1λ2)2

(λ21 + λ2

2)λ2=

√

λ41 + λ2

2λ21

(λ21 + λ2

2)2

=

√

λ21

λ21 + λ2

2

→ 1 as λ1 → ∞. (1.8)

If b is a fixed biinvariant metric on G and l1 is a positive real number, thenwe let gl1 denote the metric we obtain on M via the Riemannian submersionqG×M : G × M −→ M when the metric on the G–factor in G × M is l21b.When G = S3 , b will always be the unit metric. As pointed out in (1.8), gl1

converges to the original metric, g , as l1 → ∞. To emphasize this point wewill also denote g by g∞ .

Let TOG denote the tangent distribution in M to the orbits of G. Then anyplane P tangent to M can be written as

P = span{z + ka, ζ + kb}, (1.9)



where z, ζ ∈ HOGand ka, kb ∈ TOG . We let P denote the plane in T (G ×

M) that is horizontal with respect to qG×M and satisfies dp2(P ) = P wherep2: G × M −→ M denotes the projection onto the second factor. If ξ ∈ TM ,then ξ ∈ T (G × M) has the analogous relationship to ξ as P has to P . Aspointed out in [8] we have the following result.

Proposition 1.10 Let λa1 and λa

2 denote the lengths of the killing fields onG and M corresponding to ka via the procedures described in (1.3) and (1.4).Let λb

1 and λb2 have the analogous meaning with respect to kb .

(i) If the curvature of P is positive with respect to g∞ , then the curvatureof

dqG×M (P ) = span{ z +λa2

1 + λa22

λa21

ka, ζ +λb2

1 + λb22

λb21

kb } (1.11)

is positive with respect gl1 .

(ii) The curvature of dqG×M(P ) is positive with respect to gl1 if the A–tensor,AqG×M , of qG×M is nonzero on P .

(iii) If G = S3 , then the curvature of dqG×M (P ) is positive if the projectionof P onto TOG is nondegenerate.

(iv) If the curvature of P is 0 and AqG×M vanishes on P , then the curvatureof dqG×M(P ) is 0.

Proof (ii) and (iv) are corollaries of O’Neill’s horizontal curvature equation.

To prove (i) notice that the curvature of dqG×M (P ) is positive if the curvatureof its horizontal lift, P , is positive. The curvature of P is positive if its image,P , under dp2 has positive curvature, proving (i).

Let p1: G × M −→ G be the projection onto the first factor. The curvature ofP is also positive if its image under dp1 is positively curved. If G = S3 , thenthis is the case, provided the image of p is nondegenerate, proving (iii).

Using (1.8) we get the following.

Proposition 1.12 Let AH : H ×M −→ M be an action that is by isometrieswith respect to both g∞ and gl1 . Let HAH

denote the distribution of vectorsthat are perpendicular to the orbits of AH .

P is in HAHwith respect to g∞ if and only if dqG×M (P ) is in HAH

withrespect to gl1 .



Proof Just combine our description of gl1 with the observation that if wesquare the expression in (1.8) we get the reciprocal of the quantity in (1.11).

Ultimately we will be studying Cheeger Perturbations via commuting groupactions, AG1

, AG2, that individually have property (1.5). Generalizing our for-

mulas to this situation is a simple matter once we observe the following result.

Proposition 1.13 Let G1 × G2 act isometrically on (M,g). Fix biinvariantmetrics b1 and b2 , on G1 and G2 . Let g1 and g2 be the metrics obtained bydoing Cheeger perturbations of (M,g) with G1 and G2 respectively.

(i) G2 acts by isometries on (M,g1) and G1 acts by isometries on (M,g2).

(ii) Let g1,2 denote the metric obtained by doing the Cheeger perturbationwith G2 on (M,g1) and let g2,1 denote the metric obtained by doing theCheeger perturbation with G1 on (M,g2). Then

g1,2 = g2,1.

In fact g1,2 coincides with the metric obtained by doing a single Cheegerperturbation of (M,g) with G1 × G2 .

(iii) (k1u, k2

u, u) is horizontal for G1 × G2 × MqG1×G2×M−→ M with respect to

b1 × b2 × g if and only if (k1u, u) is horizontal for G1 × M

qG1×M−→ M with

respect to b1 × g and (k2u, u) is horizontal for G2 × M

qG2×M−→ M withrespect to b2 × g .

Proof The proof of (i) is a routine exercise in the definitions which we leaveto the reader.

Part (i) gives us a commutative diagram

G1 × G2 × Mid × qG1×M−−−−−−−−→ G2 × M

id × qG2×M

y

y

qG2×M

G1 × MqG1×M−−−−→ M

of Riemannian submersions from which (ii) readily follows.

It follows from the diagram that if (κ1, κ2, u) is horizontal for qG1×G2×M withrespect to b1 × b2 × g , then (κ1, u) is horizontal for qG1×M with respect tob1 × g and (κ2, u) is horizontal for qG2×M with respect to b2 × g . This provesthe “only if” part of (iii).

The “if” part of (iii) follows from the “only if” part and the observation thatHqG1×G2×M

∩ T (G1 × G2) = 0 via a dimension counting argument.



Rather than changing the metric of E2,0 directly with a Cheeger perturbation,we will change the metric on Sp(2) and then mod out by A2,0 . The constraintto this approach is that the Cheeger perturbations that we use can not destroythe fact that A2,0 is by isometries.

Fortunately it was observed in [8] that if the metric on the G factor in G×Mis biinvariant, then G acts by isometries with respect to gl1 . Therefore we havethe following result.

Proposition 1.14 Let gν1,ν2,lu1,ld1

denote a metric obtained from the biinvari-

ant metric on Sp(2) via Cheeger’s method using the S3 ×S3 ×S3 ×S3–action,Au × Ad × Al × Ar .

Then Au×Ad×Al×Ar is by isometries with respect to gν1,ν2,lu1,ld1

. In particular,A2,0 is by isometries with respect to gν1,ν2,lu

1,ld1

.

We include a proof of Proposition 1.14 even though it follows from an assertionon page 624 of [8]. We do this to establish notation that will be used in thesequel, and because the assertion in [8] was not proven.

Proof Throughout the paper we will call the tangent spaces to the orbits ofAl and Ar , V1 and V2 . The orthogonal complement of V1 ⊕ V2 with respectto the biinvariant metric will be called H . According to Proposition 2.1 in

[24], Sp(2) is diffeomorphic to the pull back of the Hopf fibration S7 h−→ S4

via S7 a◦h−→ S4 , where a: S4 −→ S4 is the antipodal map and S7 h−→ S4 isthe Hopf fibration that is given by right multiplication by S3 . Moreover, themetric induced on the pull back by the product of two unit S7 ’s is biinvariant.Through out the paper our computations will be based on perturbations of thebiinvariant metric, b 1√

2

, induced by S7( 1√2) × S7( 1√

2), where S7( 1√

2) is the

sphere of radius 1√2.

Observe that if k is a killing field on S3 whose length is 1 with respect tothe unit metric, then the corresponding killing field on Sp(2) with respect toeither Al or Ar has length 1√

2with respect to b 1√

2

. It follows from this that

the quantity (1.8) is constant when we do a Cheeger perturbation on b 1√2

via

either Al or Ar . Thus the effect of these Cheeger perturbations is to scale V1

and V2 , and to preserve the splitting V1 ⊕ V2 ⊕ H and b 1√2

|H . The amount

of the scaling is < 1 and converges to 1 as the scale, l1 , on the S3–factor inS3 × Sp(2) converges to ∞ and converges to 0 as l1 → 0. We will call the



resulting scales on V1 and V2 , ν1 and ν2 , and call the resulting metric gν1,ν2.

With this convention the biinvariant metric b 1√2

is g 1√2, 1√

2

.

It follows that gν1,ν2is the restriction to Sp(2) of the product metric S7

ν1×S7

ν2

where S7ν denotes the metric obtained from S7( 1√

2) by scaling the fibers of h

by ν√

2. Since Al and Ar are by symmetries of h in each column, they are byisometries on S7

ν1× S7

ν2and hence also on (Sp(2), gν1,ν2

).

Let glu1,ld1

denote the metric obtained from b 1√2

via the Cheeger perturbation

with Au × Ad when the metric on the S3 × S3–factor in S3 × S3 × Sp(2) isS3(lu1 ) × S3(ld1). An argument similar to the one above, using rows instead ofcolumns, shows that Au × Ad is by isometries on (Sp(2), glu

1,ld1

).

Since Au×Ad commutes with Al×Ar , it follows that Au×Ad acts by isometrieson (Sp(2), gν1,ν2

). Doing a Cheeger perturbation with Au×Ad on (Sp(2), gν1,ν2)

produces a metric gν1,ν2,lu1,ld1

which can also be thought of as obtained from

Sp(2) via a single Cheeger perturbation with Al ×Ar ×Au ×Ad . Since Al ×Ar

acts by isometries on (Sp(2), gν1,ν2) and commutes with Au ×Ad , Al ×Ar acts

by isometries with respect to gν1,ν2,lu1,ld1

.

But gν1,ν2,lu1,ld1

can also be obtained by first perturbing with Au × Ad and

then perturbing with Al × Ar . So repeating the argument of the proceedingparagraph shows that Au×Ad acts by isometries with respect to gν1,ν2,lu

1,ld1

.

It follows that q2,0: Sp(2) −→ E2,0 = Sp(2)/A2,0 is a Riemannian submersionwith respect to both gν1,ν2

and gν1,ν2,lu1,ld1

. We will abuse notation and call theinduced metrics on E2,0 , gν1,ν2

and gν1,ν2,lu1,ld1

.

2 Symmetries and their effects on E2,0

Since Al × Ar commutes with A2,0 and is by isometries on (Sp(2), gν1 ,ν2,lu1,ld1

),

it is by isometries on (E2,0, gν1,ν2,lu1,ld1

). However on the level of E2,0 it hasa kernel that at least contains Z2 . To see this just observe that the action of(−1,−1) on Sp(2) via Al×Ar is the same as the action of −1 via A2,0 . It turnsout that the kernel is exactly Z2 , and that Al × Ar induces the SO(4)–actionwhose existence was asserted in Theorem A(i).

Proposition 2.1 (i) Al and Ar act freely on E2,0 .



(ii) E2,0/Al is diffeomorphic to S4 and the quotient map p2,0: E2,0 −→

E2,0/Al is the bundle of type (2, 0).

(iii) Ar acts by symmetries of p2,0 . The induced map on S4 is the join of thestandard Z2–ineffective S3–action on S2 with the trivial action on S1 .

(iv) The kernel of the action Al × Ar on E2,0 is generated by (−1,−1), soAl × Ar induces an effective SO(4)–action on E2,0 .

Proof It is easy to see that A2,0 × Al and A2,0 × Ar are free on Sp(2) andhence that Al and Ar are free on E2,0 , proving (i).

It was observed in [12] that Sp(2)/(A2,0 ×Al) = E2,0/Al is diffeomorphic to S4

and in [20] that p2,0 is the bundle of type (2, 0). Combining these facts proves(ii).

Part (iii) is a special case of Proposition 5.5 in [24].

It follows from (iii) that if (q, p) is in the kernel of Al × Ar , then p = −1.

On the other hand, Al is the principal S3–action for E2,0p2,0−→ S4 . Combining

these facts we see that the kernel has order ≤ 2, and we observed above that(−1,−1) is in the kernel.

The O(2) action on Sp(2), AO(2): O(2) × Sp(2) −→ S(2), that is given by

AO(2): (A,U) 7→ AU

commutes with A2,0 and so is by symmetries of q2,0: Sp(2) −→ E2,0 . Sinceit also commutes with Al × Ar it acts by isometries on (E2,0, gν1,ν2

). It isalso acts by symmetries of p2,0 according to Proposition 5.5 in [24]. We shallsee, however, that it is not by isometries on (Sp(2), gν1 ,ν2,lu

1,ld1

). (Note that it

commutes with neither Au nor Ad . This is one of the central ideas behind thecurvature computations in the proof of Theorem A.)

We may nevertheless use AO(2) to find the 0–curvatures of (E2,0, gν1,ν2). (We

will then use Proposition 1.10 to see that most of them are not 0 with respectto gν1,ν2,lu

1,ld1

.)

A special case of Proposition 5.7 in [24] is the following.

Proposition 2.2 Every point in E2,0 has a point in its orbit under ASO(2) ×Al × Ar that can be represented in Sp(2) by a point of the form

( (

cos tα sin t

)

,

(

α sin tcos t

) )

, (2.3)

with t ∈ [0, π4 ], re(α) = 0 and |α| = 1.



We will call points of the form (2.3) representative points.

Notational Convention We have seen that Ar , Al and ASO(2) all induceactions on E2,0 , and that Ar and ASO(2) even induce actions on S4 . To simplifythe exposition we will make no notational distinction between these actions andtheir induced actions. Thus Ar stands for an action on Sp(2), E2,0 , or S4 . Thespace that is acted on will be clear from the context.

3 The curvature tensor of (Sp(2), gν1,ν2)

We will study the curvature of (Sp(2), gν1,ν2) by analyzing the geometry of the

riemannian submersion p2,1: Sp(2) −→ S7 given by projection onto the firstcolumn.

The moral of our story is that the geometry of p2,1 resembles the geometry ofh: S7 −→ S4 very closely. p2,1 is a principal bundle with a “connection metric”.That is the metric is of the form

〈X, Y 〉t,ω = 〈dpE(X), dpE(Y )〉B + t2〈ω(X), ω(Y )〉G,

where G → EpE−→ B is a principal bundle, 〈 , 〉G is a biinvariant metric on G,

〈 , 〉B is an arbitrary metric on B , and ω: V E −→ TE is a connection map.In the case of p2,1 , G = S3 and the action is given by Ar .

It will be important for us to understand how the infinitesimal geometry ofp2,1 changes as ν1 and ν2 change. Combining 2.1–2.3 of [17] with a rescalingargument we get the following.

Proposition 3.1 Let G → EpE−→ B be a principal bundle with a connection

metric 〈 , 〉t,ω . Let ∇t , At and Rt denote the covariant derivative, A–tensorand curvature tensor of 〈 , 〉t,ω . If the paremeter t is omitted from the su-perscript of one of the objects ∇t , At , Rt , or 〈 , 〉t,ω , then implicity thatparameter has value 1.

If e1 , e2 and Z are horizontal fields and σ , U and τ are vertical fields, then:

(i) The fibers of pE are totally geodesic.

(ii) Ate1

e2 = Ae1e2 , At

e1σ = t2Ae1

σ , ∇te1

e2 = ∇e1e2 , (∇t

e1σ)v = (∇e1

σ)v ,(∇t

στ)v = (∇στ)v ,

(iii) 〈Rt(e1, e2)e2, e1〉t = 〈RB(dpE(e1), dpE(e2))dpE(e2), dpE(e1)〉− 3t2‖Ae1

e2‖2 ,



(iv) 〈Rt(σ, τ)τ, σ〉t = t2〈R(σ, τ)τ, σ〉,(v) 〈Rt(e1, σ)σ, e1〉t = t4‖Ae1

σ‖2 ,

(vi) 〈Rt(σ, τ)U, e1〉t = 0, and

(vii) 〈Rt(e1, e2)Z, σ〉t = t2〈R(e1, e2)Z, σ〉.

The next step is to determine the A and T tensors of p2,1 .

The vertical space of p2,1 is

Vp2,1= { (0, σ) ∈ TSp(2) | σ ∈ Vh} ≡ V2,

where Vh is the vertical space for the Hopf fibration h: S7 −→ S4 that is givenby left quaternionic multiplication. We will also denote vectors in Vp2,1

by(0, τ), and we will often abuse notation and write just τ for (0, τ).

The T–tensor Given two vector fields, (0, σ1), (0, σ2), with values in Vp2,1

we computeT

p2,1

(0,σ1)(0, σ2) = (0,∇σ1σ2)

h = (0, 0)

since Vh is totally geodesic. Therefore

T ≡ 0. (3.2)

The A–tensor Hp2,1splits as the direct sum Hp2,1

= V1 ⊕ H where V1 andH are as defined on page 339. So any vector field with values in Hp2,1

can bewritten uniquely as z +w where z takes values in H and w takes values in V1 .Given two such vector fields z1 + w1 and z2 + w2 we compute

Ap2,1

z1+w1z2 + w2 = (∇z1+w1

z2 + w2)v = (∇z1

z2)v =

( 0, ( ∇S7ν2

dp2

2(z1)

dp22(z2) )v) = ( 0,

1

2[dp2

2(z1), dp22(z2)]

v ) = (3.3)

( 0, ∇1dp2

2(z1)dp2

2(z2) ),

where p22: Sp(2) −→ S7 denotes the projection of Sp(2) onto its last factor.

The second equality is due to the fact that a vector in V2 is 0 in the first entryand a vector in V1 is 0 in the last entry.

The upshot of (3.2) and (3.3) is that the T and A tensors of p2,1 are essentiallythe “T and A tensors of h in the last factor”. The only difference is that Ap2,1

has a 3–dimensional kernel, V1 , and Ah has kernel = 0.

This principal also holds for the vertizontal A–tensor. If σ = (0, σ) is a vectorfield with values in V2 and z+w is a vector field with values in Hp2,1

= H⊕V1 ,then

Ap2,1

z+wσ = ( 0, ∇S7ν2

dp2

2(z)

σ)h = 2ν22( 0,∇S7(1)

dp2

2(z)

σ )h. (3.4)



The only components of the curvature tensor of Sp(2) which are not mentionedin (3.1) are those of the form 〈R(e1, σ)e2, τ〉 and 〈R(σ, τ)e1, e2〉. According toformulas {2} and {2′} in [18] these are

〈R(e1, σ)e2, τ〉 = −〈(∇σA)e1e2, τ〉 − 〈Ae1

σ,Ae2τ〉

and

〈R(σ, τ)e1, e2〉 =

−〈(∇σA)e1e2, τ〉 − 〈Ae1

σ,Ae2τ〉 + 〈(∇τA)e1

e2, σ〉 + 〈Ae1τ,Ae2

σ〉. (3.5)

(We use the opposite sign convention for the curvature tensor than O’Neill.)

If we choose e1 and e2 to be basic horizontal fields the first equation simplifiesto

〈R(e1, σ)e2, τ〉 =

−〈∇σ(Ae1e2), τ〉 + 〈A∇σe1

e2, τ〉 + 〈Ae1∇t

σe2, τ〉 − 〈Ae1σ,Ae2

τ〉 =

−〈∇σ(Ae1e2), τ〉 − 〈Ae2

(∇e1σ)h, τ〉 + 〈Ae1

(∇e2σ)h, τ〉 − 〈Ae1

σ,Ae2τ〉 = (3.6)

−〈∇σ(Ae1e2), τ〉 + 〈Ae1

σ,Ae2τ〉 − 〈Ae2

σ,Ae1τ〉 − 〈Ae1

σ,Ae2τ〉 =

−〈∇σ(Ae1e2), τ〉 − 〈Ae2

σ,Ae1τ〉,

where the superscript h denotes the component in H . It appears in the formulabecause the orthogonal complement of H in the horizontal space for p2,1 , V1 ,is the kernel of A.

Similarly (3.5) simplifies to

〈R(σ, τ)e1, e2〉 =

−〈∇σ(Ae1e2), τ〉 − 〈Ae2

σ,Ae1τ〉 + 〈∇τ (Ae1

e2), σ〉 + 〈Ae2τ,Ae1

σ〉. (3.7)

According to O’Neill, the curvatures with exactly three horizontal terms are

〈R(e1, e2)e3, σ〉 = −〈(∇e3A)e1

e2, σ〉 =

−〈∇e3(Ae1

e2), σ〉 + 〈A(∇e3e1)he2, σ〉 + 〈Ae1

(∇e3e2)

h, σ〉. (3.8)

Through out the rest of this section we will let the superscript v denote thecomponent in V1 .

By (3.3) and the fact that the analogous equation holds for the Hopf fibrationwe have

〈R(eh1 , eh

2 )eh3 , σ〉 =

−〈∇eh3

(Aeh1

eh2), σ〉 + 〈A(∇

eh3

eh1)heh

2 , σ〉 + 〈Aeh1

(∇eh3

eh2)h, σ〉 = 0. (3.9)



Of course

〈R(ev1, e

v2)e3, σ〉 = 0 (3.10)

since V1 is the kernel of A. Using this again we get

〈R(eh1 , ev

2)ev3, σ〉 = 〈Aeh

1

( ∇ev3ev2 )h, σ〉 = 0, (3.11)

where the last equality is because ( ∇ev3ev2 )h = 0, since the orbits of Al are

totally geodesic.

Also

〈R(eh1 , ev

2)eh3 , σ〉 =

−〈∇eh3

(Aeh1

ev2), σ〉 + 〈A(∇

eh3

eh1)hev

2, σ〉 + 〈Aeh1

(∇eh3

ev2)

h, σ〉 = (3.12)

〈Aeh1

(∇eh3

ev2)

h, σ〉 = −〈(∇eh3

ev2)

h, Aeh1

σ〉.Keeping in mind that H is the horizontal space for h ◦ p2,1 and that V1 is inthe vertical space for h ◦ p2,1 we can rewrite the right hand side of (3.12) andget

〈R(eh1 , ev

2)eh3 , σ〉 = −〈Ah◦p2,1

eh3

ev2, Aeh

1

σ〉. (3.13)

Using (3.13) and the antisymmetry of the curvature tensor we get

〈R(ev1, e

h2)eh

3 , σ〉 = 〈Ah◦p2,1

eh3

ev1, Aeh

2

σ〉. (3.14)

Combining the first Bianchi identity with (3.13) and (3.14) we find that

〈R(eh1 , eh

2 )ev3, σ〉 = −〈R(ev

3, eh1)eh

2 , σ〉 − 〈R(eh2 , ev

3)eh1 , σ〉 =

−〈Ah◦p2,1

eh2

ev3, Aeh

1

σ〉 + 〈Ah◦p2,1

eh1

ev3, Aeh

2

σ〉. (3.15)

It turns out that the sum of the two terms on the right is always 0 and hencethat

〈R(eh1 , eh

2 )ev3, σ〉 = 0. (3.16)

This can be seen by direct computation combining Proposition 4.1 (below) withsection 6 of [24], but there are many details to check. A less direct, but muchquicker proof begins by using the second equation in 3.1(ii) to observe that theright hand side of (3.15) is 4ν2

1ν22 times the corresponding curvature quantity for

the biinvariant metric. Therefore it suffices to prove (3.16) for the biinvariantmetric. To do this observe that distributions V1 and V2 have the form

V1,Q =

{

LQ,∗

(

β 00 0

) ∣

∣

∣

∣

β ∈ H and re(β) = 0

}



V2,Q =

{

LQ,∗

(

0 00 β

)∣

∣

∣

∣

β ∈ H and re(β) = 0

}

,

where LQ,∗ denotes the differential of left translation by Q.

Thus [Ev3 ,Σ] = 0, where Ev

3 and Σ denote the extensions of ev3 and σ to left

invariant fields. Therefore

〈R(eh1 , eh

2)ev3, σ〉 = −1

4〈[Eh

1 , Eh2 ], [Ev

3 ,Σ]〉 = 0,

where Eh1 and Eh

2 denote the extensions of eh1 and eh

2 to left invariant fields,proving (3.16).

Combining (3.10), (3.11), and (3.16) gives us

〈R(e1, e2)ev3, σ〉 = 0. (3.17)

A review of the results of this section shows that the only difference betweenthe infinitesimal geometries of h and p2,1 is the fact that for p2,1 the terms(3.13) and (3.14) are not always 0 but for h the curvatures terms with exactlythree horizontal vectors are always 0. This is the root cause of many of the 0curvatures in (Sp(2), gν1 ,ν2

).

4 The A–tensor of h

In section 3 we showed that the A–tensor of p2,1 is essentially the A–tensorof the Hopf fibration, h, in the last factor. Ah certainly ought to be very wellknown, but the authors are not aware of any computation of it in the literature.So for completeness we prove the following.

Proposition 4.1 Let N be a point in S7 .

(i) The vertical space of h at N , Vh,N , is

{Nβ | β ∈ H, re(β) = 0}.(ii) If z is in the horizontal space, Hh,N , for h at N , then

AhzNβ = zβ.

Proof A proof of this for the case of the complex Hopf fibration can be foundin [18]. The proof for the quaternionic case is nearly identical.

(i) is an immediate consequence of the definition h.



To prove (ii) we let a superscript h denote the horizontal component withrespect to h and we compute

(∇S7

z Nβ)h = (∇R8

z Nβ)h = ((∇R8

z N)β)h = (zβ)h = zβ,

where for the last equality we have used the fact that Hh,N is invariant underright quaternionic multiplication.

5 The Curvature of Generic Planes in Sp(2)

Because the metric gν1,ν2is a simpler metric than gν1,ν2lu

1,ld1

, we study its 0

curvatures in this section. Where all statements about curvatures of Sp(2) areunderstood to be with respect to gν1,ν2

, for some ν1, ν2 < 1√2.

A plane P tangent to Sp(2) has the form

P = span{z + w, ζ + v}, (5.1)

where z, ζ ∈ H and w, v ∈ V1 ⊕ V2 .

Theorem 5.2

(i) The curvature of P is positive if the projection of P on to one of H , V1 ,or V2 is two dimensional.

(ii) If all three projections in (i) are degenerate, then the curvature of P is0, if its projection onto H is 0.

(iii) If all three projections in (i) are degenerate and the projection of P ontoH is one dimensional, then we may assume that z 6= 0 and ζ = 0. Thecurvature of P is positive unless

Ah◦p2,1z v1 + A

p1,2z v2 = 0. (5.3)

(iv) If all of the hypotheses of (iii) hold and (5.3) holds, then the curvature ofP is 0.

Remark Since v2 is vertical for h ◦ p1,2 and Ap1,2z v2 = A

h◦p1,2z v2 , (5.3) can

be thought of as Ah◦p1,2z (v1 + v2) = 0. Since the fibers of h ◦ p1,2 are totally

geodesic this version of (iii) is a corollary of 3.1(v). We give an alternative proofbelow.



Proof of (iii) and (iv) If the projection of P onto H , V1 and V2 is degener-ate, then by replacing the second vector with the appropriate linear combinationof the vectors in (5.1) we may assume that P has the form

P = span{z + w1 + w2, v1 + v2}, (5.4)

where w1, v1 ∈ V1 and w2, v2 ∈ V2 are multiples of each other.

By replacing the first vector in (5.4) by the appropriate vector in P we canfurther assume that w2 = 0, so P has the form

P = span{z + w1, v1 + v2},where w1 and v1 are multiples of each other.

It follows that

〈 R(z + w1, v1 + v2) v1 + v2, z + w1 〉 =

〈 R(z, v1) v1, z 〉 + 2〈 R(z, v1) v2, z 〉 + 〈 R(z, v2) v2, z 〉 =

‖ Ah◦p2,1z v1 ‖2 + 2〈 A

h◦p2,1z v1, A

p1,2z v2〉 + ‖ A

p1,2z v2 ‖2 = (5.5)

‖ Ah◦p2,1z v1 + A

p1,2z v2 ‖2,

where we have used the results of section 3 to conclude that many terms are 0.

(iii) and (iv) follow from (5.5).

Notation Since the A–tensors in (5.5) are essentially the A–tensors of h inthe first and second factor we will shorten the notation and set

A1 = Ah◦p2,1 and

A2 = Ap1,2,

and we will denote the inner product in (5.5) by

〈 Ah◦p2,1z v1, A

p1,2z v2〉 = 〈 A1

zv1, A2

zv2〉a.

The idea behind this notation is that we are taking a certain type of innerproduct between the A–tensors of h in the first and second factors. The sub-script a is meant to remind us of the role that the antipodal map of S4 playsin determining this inner product.

Proof of (i) If the projection of P onto either V1 or V2 is nondegenerate,then according to 1.10(iii) the curvature of P is positive.

So we may assume that the projection of P onto H is nondegenerate and theprojections onto both V1 and V2 are degenerate. Since the projection of P onto



V2 is degenerate we may replace the first vector with another vector in P thatsatisfies

w2 = 0. (5.6)

Since the projection onto V1 is also degenerate we may replace our vectors byappropriate linear combinations to get that either

v1 = 0 or w1 = 0. (5.7)

If v1 = 0, P is spanned by

{z + w1, ζ + v2}.It follows that

〈R(z + w1, ζ + v2)ζ + v2, z + w1〉gν1,ν2≥

|z|2|ζ⊥|2 + |A1ζw

1|2 + |A2zv

2|2 +

〈R(z, ζ)v2, w1〉 + 〈R(z, v2)ζ, w1〉 + 〈R(w1, v2)ζ, z〉 + 〈R(w1, ζ)v2, z〉 +

〈R(z, v2)v2, w1〉 + 〈R(w1, v2)v2, z〉,where ζ⊥ denotes the component of ζ that is perpendicular to z , and we haverelied heavily on results of the previous two sections to conclude that many ofthe components of this curvature are 0.

Using (3.17) and the symmetries of the curvature tensor we see that the fourthand the sixth terms on the right are 0. The last two terms are also 0. One wayto see this is to combine (3.6) with the fact that w1 is in the kernel of Ap2,1 .

Use (3.13) to evaluate the other two terms to get

〈R(z + w1, ζ + v2)ζ + v2, z + w1〉gν1,ν2≥

|z|2|ζ⊥|2 + |A1ζw

1|2 + |A2zv

2|2 − 2〈A1zw

1, Aζv2〉a. (5.8)

For ξ ∈ TSp(2) let |ξ|1,1 = |dp2,1ξ|1 and |ξ|1,2 = |dp22ξ|1 where | · |1 denotes the

unit norm on S7 . Combining 3.1(v), (3.4) and (4.1) and meticulously chasingthrough the definitions we get that

|A1ζw

1|2 = ν41 |ζ|21,1 |w1|21,1,

|A2zv

2|2 = ν42 |z|21,2 |v2|21,2, and (5.9)

| 2〈A1zw

1, Aζv2〉a | ≤ 2ν2

1 ν22 |ζ|1,1 |w1|1,1 |z|1,2 |v2|1,2.

From (5.9) we see that sum of the last three terms in (5.8) is always nonnegativeso

〈R(z + w1, ζ + v2)ζ + v2, z + w1〉gν1,ν2≥ |z|2|ζ⊥|2 > 0, (5.10)



where the last inequality is due to our hypothesis that the projection of P ontoH is nondegenerate.

It remains to consider the case w1 = 0 in (5.7). If so, our plane is spanned by

{z, ζ + v1 + v2}.Arguing as in (5.8) and (5.10) we can show that

〈R(z, ζ + v1 + v2)ζ + v1 + v2, z〉 ≥|z|2|ζ⊥|2 + |A1

zv1|2 + |A2

zv2|2 + 2〈A1

zv1, A2

zv2〉a ≥

|z|2|ζ⊥|2 > 0.

Proof of (ii) If P = span{v1, w2} where v1 ∈ V1 and w2 ∈ V2 , then curv(P )= 0 because the T –tensor of p2,1 is 0 and V1 is the kernel of the A–tensor ofp2,1 .

6 Horizontal space of q2,0

From this point forward we will use the notation from section 6 in [24] for specifictangent vectors to Sp(2). Recall that t ∈ [0, π

4 ] is one of the coordinates of ourrepresentative points (see Proposition 2.3).

Proposition 6.1 For t ∈ [0, π4 ) the horizontal space of q2,0 with respect to

gν1,ν2is spanned by

{ (x, x), (−y, y), (η1, η1 + tan(2t)ϑ1

ν22

), (η2, η2 + tan(2t)ϑ2

ν22

),

(− v

ν21

,v

ν22

), (−ϑ1

ν21

,ϑ1

ν22

), (−ϑ2

ν21

,ϑ2

ν22

) }.

Notation We will call the seven vectors in (6.1) x2,0, y2,0, η2,01 , η2,0

2 , v2,0 ,

ϑ2,01 , and ϑ2,0

2 respectively. We will call the set consisting of the first fourbasis(H2,0), and the set consisting of the last three basis(V2,0).

Sketch of Proof The proof is just straight forward computations of innerproducts. Many of the required computations were done in Proposition 6.5 of[24].

A corollary of 5.2(i) is the following.



Corollary 6.2 Let P be a plane in E2,0 whose t–coordinate is in [0, π4 ).

If the horizontal lift of P with respect to gν1,ν2has a nondegenerate projection

onto either H2,0 or V2,0 , then P is positively curved with respect to gν1,ν2.

Proof We will combine Theorem 5.2(i) with the following observations.

(a) Any plane in H2,0 has a nondegenerate projection (with respect to thesplitting H ⊕ V1 ⊕ V2 ) onto H .

(a ′) V2,0 ⊂ V1 ⊕ V2 .

It follows from (a) and (a ′ ) that if the projection of P onto H2,0 is nondegener-ate, then so is the projection onto H (with respect to the splitting H⊕V1⊕V2).

(b) Any plane in V2,0 has a nondegenerate projection (with respect to thesplitting H ⊕ V1 ⊕ V2 ) onto V1 .

(b ′) H2,0 ⊂ H ⊕ V2 .

It follows from (b) and (b ′) that if the projection of P onto V2,0 is nonde-generate, then so is the projection of P onto V1 (with respect to the splittingH ⊕ V1 ⊕ V2 ).

Proposition 6.3 For t = π4 the horizontal space of q2,0 with respect to gν1,ν2

is spanned by

(x, x), (−y, y), ( 0,ϑ1

ν22

), ( 0,ϑ2

ν22

),

(− v

ν21

,v

ν22

), (−ϑ1

ν21

,ϑ1

ν22

), (−ϑ2

ν21

,ϑ2

ν22

).

Sketch of proof Again the proof is just straight forward computations ofinner products. In fact the computations that proved (6.1) will suffice for all ofthe vectors except ( 0, ϑ1

ν2

2

) and ( 0, ϑ2

ν2

2

). These are just multiples of the limits

ofη2,01

|η2,01

| andη2,02

|η2,02

| as t → π4 .

Notational Convention Let π: M −→ B be a riemannian submersion. Tosimplify the exposition we will make no notational distinction between a vectorthat is tangent to B and its horizontal lift to M .

Remark Since the seven vectors listed in (6.3) are limits of (function multiplesof) the seven vectors listed in (6.1) we will call them x2,0, y2,0, η2,0

1 , η2,02 ,



v2,0 , ϑ2,01 , and ϑ2,0

2 . Notice that via q2,0 ◦p2,0 , x2,0 and y2,0 project to the unitnormals of S2

im , where S2im is the 2–sphere in S4 that is fixed by the action

induced on S4 by Ar via p2,0: E2,0 −→ S4 (cf the notational remark in section6 of [24]). Because of this their definitions do not seem to be very canonicalwhen t = π

4 . The approach we have taken is to define them at representativepoints with t = π

4 as extensions of their definitions at representative pointswhen t is generic. We will extend these definitions (as needed) to all pointswith t = π

4 by letting the isometry group act.

Also notice that the vectors (ϑ1

ν2

1

, 0), (ϑ2

ν2

1

, 0) are in Hq2,0when t = π

4 . Notice

furthermore that at t = π4

curvν1,ν2( (

ϑi

ν21

, 0 ), η2,0j ) = 0, (6.4)

for any ( ϑi

ν2

1

, 0 ) ∈ span{( ϑ1

ν2

1

, 0 ), ( ϑ2

ν2

1

, 0 )} and any η2,0j ∈ span{η2,0

1 , η2,02 }.

Here and in the sequel curv(v,w) stands for 〈R(v,w)w, v〉.

7 Zero Curvatures of (E2,0, gν1,ν2) and (E2,0, gν1,ν2,lu1 ,ld

1)

It follows from (6.2) that any 0–curvature planes for (E2,0, gν1,ν2) have the

form

P = span{ζ, w}, (7.1)

where ζ ∈ H2,0 , w ∈ V2,0 and t ∈ [0, π4 ). About such planes we will prove the

following.

Proposition 7.2

(i) For t ∈ [0, π4 ) the 0–curvature planes of (E2,0, gν1,ν2

) satisfy (7.1) andalso,

ζ ∈ span{x2,0, η2,0}w ∈ span{ϑ2,0}, (7.3)

where η2,0 is a convex combination of {η2,01 , η2,0

2 } and ϑ2,0 is the same

convex combination of {ϑ2,01 , ϑ2,0

2 }.

(ii) When t = π4 the planes of the form (6.4) have 0 curvature in Sp(2) with

respect to gν1,ν2, and, in addition, given any convex combination z of

{x2,0, y2,0} and any convex combination w of { ( −ϑ1

ν2

1

, 0 ), ( −ϑ2

ν2

1

, 0 ) },



there is a unique convex combination vw,z of { ( 0, ϑ1

ν2

1

), ( 0, ϑ2

ν2

1

) }, so

that

curv(Sp(2),gν1,ν2)(z + λw,w + vw,z) = 0 (7.4)

for all λ ∈ R.There are no other zero curvatures when t = π

4 .

Proof To prove (i) split ζ into ζ = z + η where z ∈ span{x2,0, y2,0} andη ∈ span{η2,0

1 , η2,02 }. To be more concrete suppose that η is a multiple of η2,0

1 .

Suppose w has the form w = (−N1β

ν2

1

, N2β

ν2

2

) for some purely imaginary unit

quaternion β . Notice that

〈A1y2,0(−

N1β

ν21

), A2y2,0

N2β

ν22

〉 = 1, (7.5)

and

〈A1y2,0(−

N1β

ν21

), A2ξ

N2β

ν22

〉 = 〈A1ξ(−

N1β

ν21

), A2y2,0

N2β

ν22

〉 = 0, (7.6)

if ξ ∈ span{x2,0, η2,01 , η2,0

2 }. Combining 5.2(iii),(iv) with (7.5) and (7.6) we getthat the curvature of P is positive except possibly if ζ ∈ span{x2,0, η}.

On the other hand, the projection of P onto V2 is nondegenerate unless η = 0or

pV2(η) and pV2

(w) are multiples of each other. (7.7)

So P is positively curved unless η = 0 or w is a multiple of the same convexcombination of {ϑ2,0

1 , ϑ2,02 } as η is of {η2,0

1 , η2,02 }.

To complete the proof of (i) it remains to consider the case when ζ = x2,0

in which it is easy to see, using 5.2(iii),(iv) that P is positively curved unlessw ∈ span{ϑ2,0

1 , ϑ2,02 }.

We saw in (6.3) that at points where t = π4 ,

Hq2,0= span{(x, x), (−y, y), ( 0,

ϑ1

ν22

), ( 0,ϑ2

ν22

),

(− v

ν21

,v

ν22

), (−ϑ1

ν21

, 0 ), (−ϑ2

ν21

, 0 )}. (7.8)

Using (7.8) it is easy to see that the only planes in Hq2,0∩ (V1 ⊕ V2) whose

projections onto both V1 and V2 are degenerate are those of the form (6.4).



It follows that the only other possible 0’s are of the type described in 5.2(iii),(iv).Combining this with straightforward computations of A–tensors completes theproof of (ii), and even supplies explicit solutions for vw,z in terms of w and z .For example when z = x2,0 and w = ( −ϑ1

ν2

1

, 0 ), vw,z = ( 0, ϑ1

ν2

2

), and when

z = y2,0 and w = ( −ϑ1

ν2

1

, 0 ), vw,z = ( 0, −ϑ1

ν2

2

).

Remark We have shown that the planes in (7.2) have 0 curvature in (Sp(2),gν1,ν2

). They also have zero curvature in (E2,0, gν1,ν2). To verify this we have

to evaluate the A–tensor of q2,0 on these planes. It turns out that it is 0. Wewill not verify this since most of these planes become positively curved withrespect to gν1,ν2,lu

1,ld1

.

Now we study the zero curvatures of (E2,0, gν1,ν2,lu1,ld1

). To do this we adopt thefollowing convention.

Notational Convention Let Hq2,0, gν1,ν2denote the horizontal space of q2,0

with respect to gν1,ν2and Hq2,0, g

ν1,ν2,lu1

,ld1

the horizontal space of q2,0 with

respect to gν1,ν2,lu1,ld1

. Let qdiag

Au×Ad : S3 × S3 × Sp(2) −→ Sp(2) be the quotient

map given in (1.1) for Au × Ad . According to (1.12) P is in Hq2,0, gν1,ν2if

and only if dqdiag

Au×Ad(P ) is in Hq2,0, gν1,ν2,lu

1,ld1

. To keep the notation simpler

we will think of this correspondence as a parameterization of Hq2,0, gν1,ν2,lu

1,ld1

by Hq2,0, gν1,ν2and we will denote vectors and planes in Hq2,0, g

ν1,ν2lu1

,ld1

by

the corresponding vectors and planes in Hq2,0, gν1,ν2. We will do this without

any further mention or change in notation. For example x2,0 now stands fordqdiag

Au×Ad(x2,0).

With this convention it follows from 1.10(i) that the zero curvatures of gν1,ν2,lu1,ld1

are a subset of the zero curvatures of gν1,ν2. We will show that most of the

planes in (7.2) have nondegenerate projections onto the orbits of Au and henceare positively curved with respect to gν1,ν2,lu

1,ld1

according to 1.10(iii). The rootcause of this is that ASO(2) is not by isometries with respect to gν1,ν2,lu

1,ld1

. This

seems extremely plausible because ASO(2) commutes with neither Au nor Ad .A rigorous proof that it does not act by isometries follows from our curvaturecomputations.

In any case we no longer know that ASO(2) is by isometries, so (2.3) are nolonger our representative points.

The representative points are now(

cos θ sin θ− sin θ cos θ

) ( (

cos tα sin t

)

,

(

α sin tcos t

) )

=



( (

cos θ cos t + α sin θ sin t− sin θ cos t + α cos θ sin t

)

,

(

α cos θ sin t + sin θ cos t−α sin θ sin t + cos θ cos t

))

≡

( N1(θ, t), N2(θ, t) ), (7.9)

for all t ∈ [0, π4 ] and for all θ ∈ [0, π). The interval for θ is [0, π) because

ASO(2) is Z2–ineffective.

Using this we will show the following.

Proposition 7.10 For any t ∈ [0, π4 ) the planes of the form

P = span{η2,0, ϑ2,0} (7.11)

have positive curvature with respect to gν1,ν2,lu1,ld1

unless θ = 0, π4 , 3π

4 , or π2 .

Here η2,0 and ϑ2,0 stand for (the same) convex combinations of {η2,01 , η2,0

2 } and

{ϑ2,01 , ϑ2,0

2 } respectively.

Proof The main idea is that most planes are eliminated because their projec-tion onto the orbits of Au are nondegenerate. Combining 1.10(iii) with (1.13)we see that it is actually enough to check that a projection is degenerate withrespect to b 1√

2

.

The tangent space to the orbit of Au is spanned by the b 1√2

–orthogonal basis

{ Uα, Uγ1, Uγ2

} ≡{ ( (

α( cos θ cos t + α sin θ sin t )0

)

,

(

α(α cos θ sin t + sin θ cos t )0

))

,

( (

γ1( cos θ cos t + α sin θ sin t )0

)

,

(

γ1( α cos θ sin t + sin θ cos t )0

))

,

( (

γ2( cos θ cos t + α sin θ sin t )0

)

,

(

γ2( α cos θ sin t + sin θ cos t )0

)) }

, (7.12)

where γ1 and γ2 are purely imaginary, unit quaternions that satisfy γ1γ2 = α.

Before we can find the projections we also need a formula for (η1, η1) along theorbits of ASO(2) . At points of the form (7.9) (η1, η1) is

(


) ( (

−γ1 sin tγ2 cos t

)

,

(

γ2 cos t−γ1 sin t

) )

=

( (

−γ1 cos θ sin t + γ2 sin θ cos tγ1 sin θ sin t + γ2 cos θ cos t

)

,

(

γ2 cos θ cos t − γ1 sin θ sin t−γ2 sin θ cos t − γ1 cos θ sin t

) )

.



To find the projections we compute

〈(η1, η1), Uγ1〉 =

⟨( (


)

,

(


) )

,

( (

γ1 cos θ cos t − γ2 sin θ sin t0

)

,

(

−γ2 cos θ sin t + γ1 sin θ cos t0

))⟩

=

1

2( − cos2 θ sin t cos t − sin2 θ cos t sin t − cos2 θ sin t cos t − sin2 θ cos t sin t ) =

− sin t cos t = −1

2sin 2t. (7.13)

According to (6.1) we should compute the inner products of ( −ϑ1

.5 , ϑ1

.5 ) and

( −ϑ2

.5 , ϑ2

.5 ) with the vectors in (7.12) with respect to b 1√2

. These are the same

as the inner products of ( −ϑ1, ϑ1 ) and ( −ϑ2, ϑ2 ) with the vectors in (7.12)with respect to b1 , where b1 = 2b 1√

2

. We will compute the latter inner products

since the notation is simpler. We will only do this explicitly for ( −ϑ1, ϑ1 ),since the computations for ( −ϑ2, ϑ2 ) are the same modulo obvious changesin notation.

〈(0, ϑ1), Uγ1〉 =

⟨( (

00

)

,

(

γ2 cos θ sin t + γ1 sin θ cos tBlah

) )

,

( (


)

,

(


))⟩

=

− cos2 θ sin2 t + sin2 θ cos2 t, (7.14)

where “Blah” stands for a nonzero, but irrelevant term.

Combining the previous 2 equations we get

〈η2,01

, Uγ1〉 = −1

2sin 2t + tan(2t)( − cos2 θ sin2 t + sin2 θ cos2 t ) (7.15)

〈Uγ1, (−ϑ1, 0)〉 =

⟨ ( (


)

,

(


))

,

( (

−γ1 cos θ cos t − γ2 sin θ sin tBlah

)

,

(

00

))⟩

=

− cos2 θ cos2 t + sin2 θ sin2 t. (7.16)

Combining (7.14) and (7.16) we get

〈(−ϑ1, ϑ1), Uγ1〉 =



− cos2 θ sin2 t + sin2 θ cos2 t − cos2 θ cos2 t + sin2 θ sin2 t =

− cos2 θ( sin2 t + cos2 t ) + sin2 θ( cos2 t + sin2 t ) =

− cos 2θ. (7.17)

〈Uγ2, ( η1, η1 )〉 =

⟨ ( (

γ2 cos θ cos t + γ1 sin θ sin t0

)

,

(

γ1 cos θ sin t + γ2 sin θ cos t0

) )

,

( (


)

,

(


) ) ⟩

=

1

2( cos2 t cos θ sin θ − sin2 t cos θ sin θ − sin2 t cos θ sin θ + cos2 t cos θ sin θ ) =

1

4sin 2θ( 2 cos2 t − 2 sin2 t ) =

1

2sin 2θ cos 2t. (7.18)

〈Uγ2, ( 0, ϑ1 )〉 =

⟨ ( (


)

,

(


) )

,

( (

00

)

,

(

γ2 cos θ sin t + γ1 sin θ cos tblah

) ) ⟩

=

2 sin θ cos θ sin t cos t =1

2sin 2θ sin 2t. (7.19)

Combining (7.18) and (7.19) we get

〈Uγ2, η2,0

1〉 =

1

2sin 2θ cos 2t +

1

2tan 2t sin 2θ sin 2t =

sin 2θ

2( cos 2t + tan2t sin 2t ). (7.20)

〈Uγ2, (−ϑ1, 0)〉 =

⟨ ( (


)

,

(


))

,

( (

−γ1 cos θ cos t − γ2 sin θ sin tBlah

)

,

(

00

))⟩

=

−2 cos θ sin θ cos t sin t = −1

2sin 2θ sin 2t. (7.21)

Combining (7.19) and (7.21) we see that

〈(−ϑ1, ϑ1 ), Uγ2〉 = 0. (7.22)



Combining (7.17), (7.20), and (7.22) we see that the projection of P onto TOAu

is nondegenerate unless cos 2θ = 0 or sin 2θ = 0, proving the proposition.

Remark One might hope to show that the planes from (7.10) with θ = 0, π4 , π

2 ,and 3π

4 are positively curved by studying the projections onto TOAd . Theseare also nondegenerate for most values of θ . Unfortunately they are degenerateprecisely when θ = 0, π

4 , π2 , and 3π

4 .

Proposition 7.23 Let η2,0 and ϑ2,0 be as in (7.10). For t ∈ [0, π4 ) if

P = span{ζ, ϑ2,0} (7.24)

where ζ ∈ span{x2,0, η2,0}, then the curvature of P is positive with respect togν1,ν2,lu

1,ld1

unless θ = 0, π4 , π

2 , or 3π4 .

Proof Along the orbit of SO(2) the formula for x2,0 is

(


) ( (

− sin tα cos t

)

,

(

α cos t− sin t

) )

=

( (

− cos θ sin t + α sin θ cos tsin θ sin t + α cos θ cos t

)

,

(

α cos θ cos t − sin θ sin t−α sin θ cos t − cos θ sin t

) )

.

So

〈Uα, x2,0〉 =⟨ ( (

α cos θ cos t − sin θ sin t0

)

,

(

− cos θ sin t + α sin θ cos t0

) )

,

( (

− cos θ sin t + α sin θ cos tsin θ sin t + α cos θ cos t

)

,

(

α cos θ cos t − sin θ sin t−α sin θ cos t − cos θ sin t

) ) ⟩

=

1

2( cos2 t sin θ cos θ + sin2 t cos θ sin θ + sin2 t cos θ sin θ + cos2 t sin θ cos θ ) =

1

2sin 2θ, (7.25)

Similar computations show that

〈Uγi, x2,0〉 = 〈Uα, η2,0

1 〉 = 〈Uα, ϑ2,01 〉 = 0. (7.26)

Combining (7.17), (7.25), and (7.26) we see that the projection of P onto TOAu

is nondegenerate unless sin 2θ = 0 or cos 2θ = 0, proving the proposition.



Remark One might hope that there really is positive curvature on these planeswhen θ = 0, π

4 , π2 and 3π

4 because perhaps the projection of these planes ontoTOAd is nondegenerate. While this is true for most values of θ , it is falseprecisely when θ = 0, π

4 , π2 and 3π

4 .

Finally we study the zero curvatures of (E2,0, gν1,ν2,lu1,ld2

) at points when t = π4 .

At such points the action induced by ASO(2) on S4 via p2,0: E2,0 −→ S4 fixesS2

im , so the orbits of ASO(2) are vertical with respect to p2,0 when t = π4 .

Combining this with the fact that Al acts transitively on the fibers of p2,0 , wesee that to study the curvature of these points we may assume that θ = 0, inother words we are back to representative points of the form (2.3) with t = π

4 .At such points we get the following as a corollary of (7.14), (7.16), (7.19) and(7.21).

Corollary 7.27 When t = π4 and θ = 0 the planes of the form (6.4) have

positive curvature except if they have the form

P = span{( ϑ, 0 ), (0, ϑ)} (7.28)

where ϑ is any convex combination of {ϑ1, ϑ2}.

To determine which of the planes in (7.4) still have 0–curvature we prove thefollowing.

Proposition 7.29 At points with t = π4 and θ = 0, the only planes of the

form (7.4) that have 0–curvature with respect to gν1,ν2,lu1,ld1

are those of theform

P = span{x2,0 + λ(−ϑ, 0), (−ϑ, ϑ)} (7.30)

and

P = span{y2,0 + λ(−ϑ, 0), (ϑ, ϑ)} (7.31)

where ϑ stands for any convex combination of {ϑ1, ϑ2} and λ is any real num-ber.

Proof The point is that if z is any other convex combination of {x2,0, y2,0},then the projection of the plane in (7.4) onto TOAu is nondegenerate. To seethis we will need to compute the projection of y2,0 onto TOAu . Although we



will only need to know its values for (t, θ) = (π4 , 0), we will do the computation

for arbitrary (t, θ), since it may be of interest to some readers.

Along the orbit of SO(2) the formula for y2,0 is(


) ( (

α sin tcos t

)

,

(

− cos t−α sin t

) )

=

( (


)

,

(

− cos θ cos t − α sin θ sin tsin θ cos t − α cos θ sin t

) )

.

So

〈Uα, y2,0〉 =⟨ ( (

α cos θ cos t − sin θ sin t0

)

,

(

− cos θ sin t + α sin θ cos t0

)

,

( (


)

,

(

− cos θ cos t − α sin θ sin tsin θ cos t − α cos θ sin t

) ) ⟩

=

1

2( cos2 θ sin t cos t − sin2 θ cos t sin t + cos2 θ sin t cos t − sin2 θ cos t sin t ) =

1

2sin t cos t(2 cos2 θ − 2 sin2 θ) =

1

2sin 2t cos 2θ. (7.32)

Similar computations show that

〈Uγi, y2,0〉 = 〈Uα, (ϑ, 0)〉 = 〈Uα, (0, ϑ)〉 = 0. (7.33)

Combining (7.32) and (7.33) with (7.14), (7.16), (7.19), (7.21), (7.25) and (7.26)we see that if P is any plane of the form of (7.4) and not of the form of either(7.30) or (7.31), then the projection of P onto TOAu is nondegenerate andhence P is positively curved.

Remark As was the case with our previous results one could study projectionsonto TOAd to prove (7.27) and (7.29). Unfortunately these projections aredegenerate on the same planes as the projections onto TOAu and hence do notgive us additional positive curvature.

The curvature computations for the proof of Theorem A are completed withthe following.

Proposition 7.34

(i) The planes in (7.11), (7.24), (7.28), (7.30) and (7.31) all have 0–curvaturein (E2,0, gν1,ν2,lu

1,ld1

).



(ii) The plane spanned by x2,0 and ϑ2,01 in E2,0 is tangent to a totally geodesic

flat 2–torus.

(iii) Let Z be the pointwise zero locus in (E2,0, gν1,ν2,lu1,ld1

), then Z is the union

of two copies of S3 × S3 that intersect along a copy of S2 × S3 .

Sketch of proof To prove (i) we must check that the A–tensors of qdiag

Au×Ad

and q2,0 vanish on these planes. These computations are rather long, but arequite straight forward so we leave them to the reader.

Recall [24] that S1R

denotes the circle in S4 that is fixed by the action inducedfrom Ar via p2,0 .

It is easy to see that x2,0 and ϑ2,01 span a 2–torus, T , in Sp(2) whose tangent

plane is horizontal with respect to q2,0 at every point. Since T is horizontal,q2,0|T is a covering map. The order of the covering is at least two because −1leaves T invariant under A2,0 . We show next that the order is exactly two.

Let cx and cϑ

2,01

be the geodesic circles that are generated by x2,0 and ϑ2,01 .

Then with respect to the join decomposition S4 = S2im ∗S1

R, p2,0(cx) is a radial

circle and for t > 0, p2,0(cϑ2,01

) is an intrinsic geodesic in a copy of S2 . Using

these observations we see that a point in p2,0(q2,0(T )) has only two preimagesunder p2,0 ◦ q2,0 from which it follows that q2,0|T is a two fold covering.

The fields x2,0 and ϑ2,01 are both parallel along T . Straightforward computation

shows that the results of transporting them from a point z ∈ T to −z agreeswith their images under the differential of the map induced on T by −1 viaA2,0 . Therefore q2,0|T is a 2–fold orientation preserving cover, and q2,0(T ) is atorus rather than a Klein bottle.

It follows from (i) that q2,0(T ) is flat and computations similar to those in theproof of (i) show that q2,0(T ) is totally geodesic.

Combining part (i) with (7.2), (7.10) and (7.23), we see that Z consists of thepoints with θ = 0, π

4 , π2 , or 3π

4 or t = π4 . Using the join structure S4 = S1

R∗S2

im

we see that the union of the points with θ = 0, θ = π2 and t = π

4 is a 3–spherein S4 . (Keep in mind that ASO(2) is Z2–ineffective on S4 .) For similar reasons

the union of the points with θ = π4 , θ = 3π

4 and t = π4 is another 3–sphere.

These two three spheres intersect along the common 2–sphere S2im and their

inverse images via p2,0 are of course diffeomorphic to S3 × S3 . Finally noticethat every point in this inverse image has zero curvatures because p2,0 is thequotient map of Ar and Ar acts by isometries.



The example, M6 , of Corollary B is

M6 = E2,0/Alβ

where β is any purely imaginary, unit quaternion, and Alβ is the circle subaction

of Al that is generated by β . When α = β the torus q2,0(T ) ⊂ E2,0 is normalto the orbit of Al

β . Some further curvature computations show that its image

in M6 is totally geodesic and flat.

8 Topological Computations

Proposition 8.1 E2,0 is diffeomorphic to the total space of the unit tangentbundle of S4 . In fact, the Riemannian submersion p1,1: E2,0 −→ S4 that isgiven by

p1,1: orbit

( (

a bc d

) )

7→ h(c, d)

is bundle isomorphic to the unit tangent bundle of S4 , where h: S7 −→ S4 isthe Hopf fibration given by left multiplication.

Proof Translating Theorem 9.5 on page 99 of [15] into our classification scheme(0.1) shows that the unit tangent bundle is the bundle of type (1, 1). We willshow via direct computations (similar to those in [12]) that (E2,0, p1,1) is alsothe bundle of type (1, 1).

As in [24] we define φ: R4 −→ R by

φ(u) =1

√

1 + |u|2,

and we define explicit bundle charts h1, h2: R4 × S3 −→ E2,0 by

h1(u, q) = orbit

(

−q quu 1

)

φ(u)

and

h2(v, r) = orbit

(

−rv r1 v

)

φ(v).



h1 and h2 are embeddings onto the open sets

U1 = orbit{(

a bc d

)

| d 6= 0 }

and

U2 = orbit{(

a bc d

)

| c 6= 0 }.

In fact their inverses are given by

h−11 ( orbit

(

a bc d

)

) = (cd

|d|2 ,−da

|d||a| )

and

h−12 ( orbit

(

a bc d

)

) = (cd

|c|2 ,cb

|c||b| ).

Thus

h−12 ◦ h1(u, q) = h−1

2 (orbit

(

−q quu 1

)

φ(u)) =

(u

|u|2 ,uqu

|u|2 ).

So (E2,0, p1,1) is the bundle of type (1, 1) and hence is the unit tangent bundleof S4 .

Next we compute the homology and first few homotopy groups of S3–bundlesover S4 to distinguish E2,0 from the known examples of 7–manifolds withpositive sectional curvature.

Proposition 8.2 Let Em,−n denote the S3–bundle over S4 of type (m,−n).

(i) The integral homology groups of Em,−n are

H0(Em,−n, Z) ∼= H7(Em,−n, Z) ∼= Z

H3(Em,−n, Z) ∼= Z

(m − n)Z, and

Hq(Em,−n, Z) ∼= {0} for all q 6= 0, 3, 7,

if m 6= n.



(ii) π1(Em,−n) ∼= π2(Em,−n) ∼= {0} and π3(Em,−n) ∼= Z

(m−n)Z

(iii) E2,0 does not have the homotopy type of any known example of a manifoldwith positive curvature.

Proof To compute the homology of Em,−n we decompose it as

Em,−n = D4 × S3 ∪gm,−nD4 × S3 (8.3)

where the gluing map gm,−n: S3 × S3 −→ S3 × S3 is given by gm,−n(u, v) =(u, umvu−n). From (8.3) and the Seifert–Van Kampen theorem it follows thatEm,−n is simply connected.

The Mayer–Vietoris sequence for the decomposition (8.3) is

· · · → Hq(S3 × S3)

Φq→ Hq(D4 × S3) ⊕ Hq(D

4 × S3)

Ψq→ Hq(Em,−n)Γq→ Hq−1(S

3 × S3)Φq−1→ · · · (8.4)

Since H2(D4 × S3) ∼= H1(S

3 × S3) ∼= {0}, H2(Em,−n) ∼= {0}.

Also since H2(S3 × S3) ∼= {0}, H3(Em,−n) ∼= H3(D4×S3)⊕H3(D4×S3)

image(Φ3) . The next

step is to compute image(Φ3).

Suppose {α×1, 1×α} is a set of generators for H3(S3×S3), and {(β, 0), (0, β)}

generates H3(D4 × S3) ⊕ H3(D

4 × S3). Then using the fact that the mapPk: S3 −→ S3 given by Pk(q) = qk has degree k we can see that

Φ3(α × 1) = (m − n)(0, β) andΦ3(1 × α) = (β, 0) + (0, β).

( To evaluate the first map represent α × 1 by S3 × {1}. )

Put another way, the matrix of Φ3 with respect to our sets of generators is(

0 1m − n 1

)

. (8.5)

From this point a routine algebraic computation shows that

H3(Em,−n) ∼= H3(D4 × S3) ⊕ H3(D

4 × S3)

image(Φ3)∼= Z

(m − n)Z.

It follows from (8.5) that Φ3 is injective if m 6= n, and hence we get using (8.4)that

H4(Em,−n) ∼= H5(Em,−n) ∼= H6(Em,−n) ∼= 0,



completing the proof of (i).

Part (ii) is a corollary of (i).

The known examples of simply connected 7–manifolds with positive curvatureare given in [3], [9] and [2]. With the exception of S7 and the example, V1 ,of Berger none of them are 2–connected. Since H3(E2,0, Z) ∼= Z2 , E2,0 isnot a homotopy sphere. It is not homotopy equivalent V1 since, according toProposition 40.1 in [3], V1 is not a Z5–cohomology sphere.

Our last topological computation is the following (cf also [13] Corollary 3.9).

Proposition 8.6

(i) M6 has the same integral cohomology modules as CP 3 but not the sameintegral cohomology algebra.

(ii) M6 does not have the homotopy type of any known example of a manifoldof positive curvature.

Proof From the long exact homotopy sequence of the fibration S1 → E2,0 −→M6 we see that π1(M

6) ∼= 0 and π2(M6) ∼= Z. Thus H1(M6, Z) ∼= 0 and

H2(M6, Z) ∼= Z. To compute H3(M6, Z) and the cup products we appeal tothe Gysin sequence

· · · → H1(M6, Z) → H3(M6, Z) → H3(E2,0, Z) → H2(M6, Z) → · · ·

of the fiber bundle S1 → E2,0 −→ M6 with integer coefficients. We alreadyknow that H1(M6, Z) = 0. Moreover, it follows from the universal coefficienttheorem that H3(E2,0, Z) = 0 since H2(E2,0, Z) has no torsion and H3(E2,0, Z)is a torsion group. Thus the Gysin sequence shows that H3(M6, Z) = 0 andhence that M6 has the same integral cohomology groups as CP 3. To see thatM6 does not have the cohomology ring of CP 3 we look at the Gysin sequenceagain:

0 = H3(E2,0, Z)π∗→ H2(M6, Z)

∪e→ H4(M6, Z)π∗

→H4(E2,0, Z)

π∗→ H3(M6, Z) = 0,

0 = H5(E2,0, Z)π∗→ H4(M6, Z)

∪e→ H6(M6, Z)π∗

→ H6(E2,0, Z) = 0.

We therefore have

0 → H2(M6, Z)∪e→ H4(M6, Z)

π∗

→ Z2 → 0,

0 → H4(M6, Z)∪e→ H6(M6, Z) → 0.



This means that H2(M6, Z)∪e→ H4(M6, Z) maps the generator x for H2(M6, Z)

to twice a generator y for H4(M6, Z). If we let e = kx we therefore havex ∪ kx = 2y. Therefore, it suffices to show that k = ±1 in order to show that

the cohomology ring of M6 is not that of CP 3. We have that H4(M6, Z)∪e→

H6(M6, Z) is an isomorphism. Thus (x ∪ kx) ∪ kx = k2x ∪ x ∪ x must alsobe twice a generator for H6(M6, Z). Therefore, if k = ±2 we would have that±2x ∪ x ∪ x generates H6(M6, Z). This, however, is impossible.

Besides S6 and CP 3 , there are two examples given in [23] and [10] of simplyconnected 6–manifolds with positive curvature. Neither of these has the coho-mology modules of CP 3 so M6 does not have the homotopy type of any knownexample.

References

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